#1
I'm getting better with wiring, but things like this are outside my knowledge still...

Is it possible to wire an LED in-line with the volume pot for a visual display of volume status?

Simple, perhaps silly but IMHO, pretty stinkin' useful
我会关闭我的耳朵,和我的心; 我会变成一个石头
"I will close my ears and my heart and I will be a stone"
#3
Nothing fancy, just fade from off to on basically.

Just to give you an idea of where your volume is without actually manipulating the pot.
我会关闭我的耳朵,和我的心; 我会变成一个石头
"I will close my ears and my heart and I will be a stone"
#5
Said guitar would probably be equipped with EMGs, but for the sake of argument, lets just say an independent 9v battery.
我会关闭我的耳朵,和我的心; 我会变成一个石头
"I will close my ears and my heart and I will be a stone"
#6
Yea, you could have a pot that controls the resistance of two sets of leads rather than the normal one. I don't know what they're called.

Anyways, you'd just hook it up like a normal LED, resistor and all, and then turning the pot would change how much electricity is passing through the LED... I don't know the specifics, but you can figure that out.

Also, I recommend drilling a hole for the LED in a place other than the actual knob itself.
TOO MANY PUPPIES

Soda sucks.
#7
Quote by Firequacker
Yea, you could have a pot that controls the resistance of two sets of leads rather than the normal one. I don't know what they're called.

Anyways, you'd just hook it up like a normal LED, resistor and all, and then turning the pot would change how much electricity is passing through the LED... I don't know the specifics, but you can figure that out.

Also, I recommend drilling a hole for the LED in a place other than the actual knob itself.


Dual gang pot. basically two pots with one shaft. but usally volume pots are quite a high valu..e.g 500k when limiting resitstors for LEDs vary around the 0-10k mark?
#8
Quote by Entrant_21
Dual gang pot. basically two pots with one shaft. but usally volume pots are quite a high valu..e.g 500k when limiting resitstors for LEDs vary around the 0-10k mark?


Is there a way to proportionately reduce the values? As in, with a component between the pot and the LED. Also, I think it would make sense to put the resistor BEFORE the pot, that way you're not just getting change at a certain point, you're getting the full range of brightness.
TOO MANY PUPPIES

Soda sucks.
#9
See, resistors and issues with values are out of my league. I've got no idea what you guys are talking about and I would probably just end up putting a 9v in-line with the pot, no resistors or anything.

Again, I have no idea how to do what I'm trying to do, hence why I asked
我会关闭我的耳朵,和我的心; 我会变成一个石头
"I will close my ears and my heart and I will be a stone"
#10
you might be better with an old fashioned torch lightbulb, it might show gradient better, make sure you get lots of spare bulbs - otherwise if it blows, you're shagged!
also, it might eat batteries.could you set it up so it disconnects the battery when you're not using it?

Perhaps a more significant problem, is putting 9V across the output of the guitar!

I think you get double pots for doing this (I don't mean stacked pots, although you could maybe glue one of those together for the same effect), but i could be wrong
#12
I'll bump it every now and again. I know it has to be possible, but how to is the hard part.

I can only assume a dual-gang pot should be able to do what I want, but the actual usage of LEDs is a mystery to me. Resistors, factoring in the value of the pots, I really have no idea how to assemble the circuit.
我会关闭我的耳朵,和我的心; 我会变成一个石头
"I will close my ears and my heart and I will be a stone"
#13
I think it'll just be battery - dual gang pot(wired as variable resistor) - led in series, won't it?

or, would you wire it so that the pot will give you a portion of the battery's voltage?

I'm assuming you mean you'll have to add other resistors and stuff to compensate for it, sos you get a nice, smooth gradient between on/off? I'd just do that by a process of trial and error - i'd get a multi meter, wire up an LED and test the resistance that give you it just on and just off, then you can work out where you need to put any more resistors on the actual circuit, no?
#14
That right there...straight up greek.
我会关闭我的耳朵,和我的心; 我会变成一个石头
"I will close my ears and my heart and I will be a stone"
#15
a dual-ganged pot is the way to go.
The resistor value to use will change depending on the LED you're using, but anywhere between 4.7k and 10k is usually used, iirc.

The problem that I see is that (as entrant_21 already said), is that a 250k or 500k pot would mean that the LED would only be lit up in the last little bit of the pot's rotation...
I suppose you could find yourself a lower-value dual-ganged pot and use a resistor across the lugs of the half you're using for volume to make it work like a 500k or 250k pot. that way the LED will be lit up for most, if not all the pots rotation, and the volume will work like normal.
then putting in a switch between the pot and battery would help keep the battery from draining when you're not using.
#16
This is way out of my league

I'm glad I brought it up though as there seems to be some interest in it.
我会关闭我的耳朵,和我的心; 我会变成一个石头
"I will close my ears and my heart and I will be a stone"
#17
Quote by james4
a dual-ganged pot is the way to go.
The resistor value to use will change depending on the LED you're using, but anywhere between 4.7k and 10k is usually used, iirc.

The problem that I see is that (as entrant_21 already said), is that a 250k or 500k pot would mean that the LED would only be lit up in the last little bit of the pot's rotation...
I suppose you could find yourself a lower-value dual-ganged pot and use a resistor across the lugs of the half you're using for volume to make it work like a 500k or 250k pot. that way the LED will be lit up for most, if not all the pots rotation, and the volume will work like normal.
then putting in a switch between the pot and battery would help keep the battery from draining when you're not using.


I think you've got it the wrong way round, i could be wrong of course...

1/Rt = 1/R1 + 1/R2


is the formula for resistors in paralell.

so you'd want to put a resistor across the LED's pot, (say the pot is 250K, when you want it to be 10k for the LED) of 10.4K, that's what i get, anyway.

you can put a 10K and a 400ohm resistor, in series, across the two lugs of the pot for the LED (middle and a side, i think)

That should give you a variable resistor of ~10K for the LED, and leave the other half of the pot for the volume alone,

Is that right?


I was thinking, you should be able to set it up so that it gives you a nice gradient all the way though turning the knob: since it's a doide it'll cut off after a certain point, you should be able to fiddle about with some resistors to compensate for this...

stick with us ,jm1681, we'll get there eventually!
Last edited by jimRH7 at May 30, 2008,
#18
You could just open the dual ganged pot and replace the carbon track with one of a lower value.
#19
Quote by jimRH7
...so you'd want to put a resistor across the LED's pot, (say the pot is 250K, when you want it to be 10k for the LED) of 10.4K, that's what i get, anyway.

you can put a 10K and a 400ohm resistor, in series, across the two lugs of the pot for the LED (middle and a side, i think)...
When you say across the lugs, you mean one end to one lug, the other to another?

I apologize, I just got good at DPDT switches, pots are a whole new world
我会关闭我的耳朵,和我的心; 我会变成一个石头
"I will close my ears and my heart and I will be a stone"
#20
yes, one end of the two resistors (wired in series, so they behave like one 10.4k resistor) would go to the middle of the three lugs, the other end would go to one of the side lugs.

when you use just the middle lug and the end lug of the pot, and leave the third lug completely alone, you turn the pot into a variable resistor.

I like mr Hankey's idea, I have virtually no experience in practical electronics - all the stuff i know is from higher tech studies and physics.

I'll draw up a diagram of what i mean, it might clear things up a little...

Edit: I just drew up a circuit daigram, I'm talking out my arse, sorry. I will return with a better idea.


Edit 2:

test ciruit:
to find the range of the LED,
(a) turn the pot till the LED is just off and no more,
(b) disconnect the the battery, and replace the LED with an ohmmeter - record resistance (I will call it Rb for short)
(c) reconnect battery, turn the pot till LED is just on and no more,
(d) disconnect the battery and replace the LED with an phmmeter - record resistance (i will call it Rd for short)
(e) record max. resistance of pot (i will call it Re for short)

Actual circuit:

R1=Re - Rd

R4=Rb

R3=the manufacturer's value of the dual gang pot

and the awkward bit:

1/R2 + 1/R3 = 1/(Rd-Rb)

rearranging gives:

R2= R3(Rd-Rb) / [R3-(Rd-Rb)]

according to me, but I might have made a mistake, if anyone reads this, could you double-check it?


I was thinking, if you don't have a 9v already in the guitar, you could use a much smaller battery, eg: AAA, this would stop it running down your pickup's battery, and would take up less space in the cavity. you could set up a spare compartment or something, so you don't have to remove the plate every time you want to replace the battery.
Last edited by jimRH7 at May 31, 2008,
#21
Quote by jimRH7
I think you've got it the wrong way round, i could be wrong of course...

1/Rt = 1/R1 + 1/R2


is the formula for resistors in paralell.

so you'd want to put a resistor across the LED's pot, (say the pot is 250K, when you want it to be 10k for the LED) of 10.4K, that's what i get, anyway.

you can put a 10K and a 400ohm resistor, in series, across the two lugs of the pot for the LED (middle and a side, i think)

That should give you a variable resistor of ~10K for the LED, and leave the other half of the pot for the volume alone,

Is that right?


I was thinking, you should be able to set it up so that it gives you a nice gradient all the way though turning the knob: since it's a doide it'll cut off after a certain point, you should be able to fiddle about with some resistors to compensate for this...

stick with us ,jm1681, we'll get there eventually!


I meant to mention that, but again this method has a downfall.. If you do put a resistor across lugs then it is permanent where the pot is still variable, so you end up tapering the pot in a style it may not have been intended for. to get a better idea of what i mean check R.G's article here http://www.geofex.com/article_folders/potsecrets/potscret.htm
the idea would work, no question but the bulbs brightness would not respond in proportion to the volume. Our ears respond logarithmically to sound, where our eyes response isnt. or maybe im just being too precise here!

e.g 1/(Parallel Resistance) = 1/R1 + 1/R2

assuming the pot is a 500k linear at 25% through it's rotation its resistance will be:
500,000/100 x 25 = 125k ohms.
at 50% it will be
500,000/100 x 50 = 250k ohms

a graph of the pots percentage rotation against its result resistance will be a straight line! i.e. 'linear'

but if we induce a parallel resistor of value 10.2k
using: 1/(Parallel Resistance) = 1/R1 + 1/R2
at 25% the resistance of the pot is 9430 ohms
at 50% the resistance is 9800 ohms

so as you can see the pot would have a steep resistance curve for a tiny portion of its full cycle.

Hope I didnt confuse anyone!
#22
Quote by Entrant_21
I meant to mention that, but again this method has a downfall.. If you do put a resistor across lugs then it is permanent where the pot is still variable, so you end up tapering the pot in a style it may not have been intended for. to get a better idea of what i mean check R.G's article here http://www.geofex.com/article_folders/potsecrets/potscret.htm
the idea would work, no question but the bulbs brightness would not respond in proportion to the volume. Our ears respond logarithmically to sound, where our eyes response isnt. or maybe im just being too precise here!

e.g 1/(Parallel Resistance) = 1/R1 + 1/R2

assuming the pot is a 500k linear at 25% through it's rotation its resistance will be:
500,000/100 x 25 = 125k ohms.
at 50% it will be
500,000/100 x 50 = 250k ohms

a graph of the pots percentage rotation against its result resistance will be a straight line! i.e. 'linear'

but if we induce a parallel resistor of value 10.2k
using: 1/(Parallel Resistance) = 1/R1 + 1/R2
at 25% the resistance of the pot is 9430 ohms
at 50% the resistance is 9800 ohms

so as you can see the pot would have a steep resistance curve for a tiny portion of its full cycle.

Hope I didnt confuse anyone!


naw, not atall, i know exactly what you mean.


That's a bit of a bastard.

Anyway to sort it? I was thinking, since volume pots are normally LOg anyway, would wiring the pot the wrong way round compensate a little for this? (i haven't read the link yet...)
#23
I just had another idea:



Bam! Visual display of pot setting: with the right taper and everything. No power source required!
#24
I reckon this thread deserves bumping - i'm anxious to see whether someone who knows more about this stuff will come on and say whether i'm wrong or not.

Those of you who disagree, why not voice your oppinions by posting in this thread?
#25
potential divider ftw?

diagram in a sec

EDIT: heres my crude perception

Get off this damn forum and play your damn guitar.
Last edited by stevo_epi_SG_wo at May 31, 2008,
#26
Quote by mr_hankey


Bam! Visual display of pot setting: with the right taper and everything. No power source required!
Simple yes, but on a dark stage it won't do you much good.
我会关闭我的耳朵,和我的心; 我会变成一个石头
"I will close my ears and my heart and I will be a stone"
#27
Quote by stevo_epi_SG_wo
potential divider ftw?

diagram in a sec

EDIT: heres my crude perception




now how could i go about getting one of these pot dividers?
google aint helping and netiher is ebay
#28
On my fuzz box I just finished building, I rigged in LED in the circuit, if I remember right, one leg went to the - part of the battery and the other went to loop 3 on the tone pot or something like that
#30
Quote by Recondite Blue
now how could i go about getting one of these pot dividers?
google aint helping and netiher is ebay


Maplin do dual gang pots, i suspect that "Radioshack"'ll sell them anaw.
#31
Quote by Recondite Blue
you wanna be more specific and help us carnyhonky?


im pretty sure I f'ed up the circuitry on this because I didnt plan this too well and my soldering sucks but that fuzz face plan, how I rigged it was that one end of the LED went to the power switch, which I attached the wire to the same led that went to #2 on the volume pot, and the other leg of the LED went to #1 of the Tone pot. I gotta go fix mine now since a wire came out when I took it apart, so maybe if I get it goin and lookin good ill take some pics and post a thread
#32
^ Very excited to see your results
我会关闭我的耳朵,和我的心; 我会变成一个石头
"I will close my ears and my heart and I will be a stone"
#33
Quote by carnyhonky
im pretty sure I f'ed up the circuitry on this because I didnt plan this too well and my soldering sucks but that fuzz face plan, how I rigged it was that one end of the LED went to the power switch, which I attached the wire to the same led that went to #2 on the volume pot, and the other leg of the LED went to #1 of the Tone pot. I gotta go fix mine now since a wire came out when I took it apart, so maybe if I get it goin and lookin good ill take some pics and post a thread


sorry but you have confused me... could you post pics of this set-up? to clarify things..
#35
Quote by jm1681
Simple yes, but on a dark stage it won't do you much good.


Find a see through version, stick a LED inside.
#36
Quote by mr_hankey
Find a see through version, stick a LED inside.


or, with a steady hand and alot of patience, You could cut out the numbers on the dial, and shine your LED through the holes.

edit: or paint the numbers with glow in the dark paint!

Could you set it up so the LED only comes on a few seconds when you turn the volume knob? To save on batteries.
#37
Okay i just remembered about this thread today and wondered about how it could be resolved..then it hit me potential divider. all this time we were thinking of thep ots use as a variable resistor when the potential divider idea is right in front of us! it seems this would be a popular mod if it works. I attached a crappy diagram and labelled the lugs and now i'll explain the purpose.

1 - Ground
2 - Output to Jack
3 - Input from switch/pickup
4 - Ground/Battery -
5 - V+ to LED
6 - V+ after current limiting resistor

If my theory is correct..which it probably isnt..this set-up should be configured as a potential divider, as opposed to a variable resistor. Its the same sort of method a volume control uses.

The dual gang pot can be any value, I would recommend 1M linear and add a 160k taper resistor across lugs 2&3 to give the pot a more logarithmic curve. this means the LED will respond linearly and the volume response will be logarithmic. Perfect!

If anyone tries this out could they let me know?

BTW the black square is the shaft of the pot
Attachments:
rough diagram.JPG